Question

B a
12. If a and B are the zeros of the quadratic polynomial p(s) = 3s2 – 6s+ 4, find the value
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1
1
+2
B
+ 3αβ.
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Answer 1

QUESTION:

• If α and β are the zeros of the quadratic polynomial p(s) = 3s^2 − 6s + 4, find the value of α/β + β/α + 2[1/α + 1/β] + 3αβ

ANSWER:

Given:

• p(s) = 3s^2 − 6s + 4
• α and β are the zeros of polynomial p(s)

To Find:

• Value of: α/β + β/α + 2(1/α + 1/β) + 3αβ

Solution:

We are given that,

⟹ p(s) = 3s^2 − 6s + 4

We know that, for a quadratic equation, ax^2 + bx + c,

⟹ Sum of zeroes = -(Coefficient of x)/(Coefficient of x^2) = -b/a

And,

⟹ Product of zeroes = (Constant)/(Coefficient of x^2) = c/a

So, in this case,

⟹ Sum of zeroes = -(-6)/3

⟹ Sum of zeroes = 6/3

⟹ Sum of zeroes(α + β) = 2

And,

⟹ Product of zeroes = (4)/3

⟹ Product of zeroes(αβ) = 4/3

Now, we need to find the value of:

⟹ α/β + β/α + 2(1/α + 1/β) + 3αβ

On simplifying,

⟹ [(α^2 + β^2)/αβ] + [2(α + β)/αβ] + 3αβ

So,

⟹ (α^2 + β^2 + 2(α + β) + 3(αβ)^2)/αβ

We know that,

⟹ a^2 + b^2 = (a + b)^2 - 2ab.

So,

⟹ [(α + β)^2 - 2αβ + 2(α + β) + 3(αβ)^2]/αβ

Now, putting the values,

⟹ [(2)^2 - 2(4/3) + 2(2) + 3(4/3)^2]/(4/3)

⟹ (4 - 8/3 + 4 + 16/3)/(4/3)

Regrouping,

⟹ (4 + 4 - 8/3 + 16/3)/(4/3)

⟹ (8 + 8/3)/(4/3)

Taking LCM in numerator,

⟹ [(24 + 8)/3]/[4/3]

⟹ (32/3)/(4/3)

3 gets cancelled,

⟹ 32/4

⟹ 8

Hence, value of α/β + β/α + 2(1/α + 1/β) + 3αβ is 8.